Problem: Simplify and expand the following expression: $ \dfrac{3}{3a - 15}+ \dfrac{3}{3a + 3}- \dfrac{4a}{a^2 - 4a - 5} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{3}{3a - 15} = \dfrac{3}{3(a - 5)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3a + 3} = \dfrac{3}{3(a + 1)}$ We can factor the quadratic in the third term: $ \dfrac{4a}{a^2 - 4a - 5} = \dfrac{4a}{(a - 5)(a + 1)}$ Now we have: $ \dfrac{3}{3(a - 5)}+ \dfrac{3}{3(a + 1)}- \dfrac{4a}{(a - 5)(a + 1)} $ The least common multiple of the denominators is: $ 9(a - 5)(a + 1)$ In order to get the first term over $9(a - 5)(a + 1)$ , multiply by $\dfrac{3(a + 1)}{3(a + 1)}$ $ \dfrac{3}{3(a - 5)} \times \dfrac{3(a + 1)}{3(a + 1)} = \dfrac{9(a + 1)}{9(a - 5)(a + 1)} $ In order to get the second term over $9(a - 5)(a + 1)$ , multiply by $\dfrac{3(a - 5)}{3(a - 5)}$ $ \dfrac{3}{3(a + 1)} \times \dfrac{3(a - 5)}{3(a - 5)} = \dfrac{9(a - 5)}{9(a - 5)(a + 1)} $ In order to get the third term over $9(a - 5)(a + 1)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{4a}{(a - 5)(a + 1)} \times \dfrac{9}{9} = \dfrac{36a}{9(a - 5)(a + 1)} $ Now we have: $ \dfrac{9(a + 1)}{9(a - 5)(a + 1)} + \dfrac{9(a - 5)}{9(a - 5)(a + 1)} - \dfrac{36a}{9(a - 5)(a + 1)} $ $ = \dfrac{ 9(a + 1) + 9(a - 5) - 36a} {9(a - 5)(a + 1)} $ Expand: $ = \dfrac{9a + 9 + 9a - 45 - 36a}{9a^2 - 36a - 45} $ $ = \dfrac{-18a - 36}{9a^2 - 36a - 45}$ Simplify: $ = \dfrac{-2a - 4}{a^2 - 4a - 5}$